Braking: regenerative or not?

[Cranium]

53 minutes ago, chriscalandro said:

Regenerative breaking probably does not produce a whole lot of power.

Of course it does.  Deceleration is just a negative acceleration.  The biggest difference is that with acceleration, you are overcoming losses due to friction on the road and air resistance. With deceleration, the friction and air resistance work for you.  But over a short period of time (ie quick speed change), these are minimal compared to the energy required to accelerate (positively or negatively).  Stopping fast is going to still have to create a lot of Joules and has to put them somewhere.  The capacitors cannot handle it (220uf caps are the biggest ones on the Ninebot).  The heat sink can't absorb energy very fast.  That leaves the battery to absorb the bulk of the energy in a short period of time.

[esaj]

1 hour ago, Cranium said:

Of course it does.  Deceleration is just a negative acceleration.  The biggest difference is that with acceleration, you are overcoming losses due to friction on the road and air resistance. With deceleration, the friction and air resistance work for you.  But over a short period of time (ie quick speed change), these are minimal compared to the energy required to accelerate (positively or negatively).  Stopping fast is going to still have to create a lot of Joules and has to put them somewhere.  The capacitors cannot handle it (220uf caps are the biggest ones on the Ninebot).  The heat sink can't absorb energy very fast.  That leaves the battery to absorb the bulk of the energy in a short period of time.

 

2 hours ago, chriscalandro said:

Regenerative breaking probably does not produce a whole lot of power. The motor controller I posted before can handle 15 amps and it is probably the size of a postage stamp * 1.5

With Vee's MCM2s & Wheelemetrics, I recorded spikes of up to 3kW during stronger braking from about 20km/h, and I'm only <60kg, plus since it wasn't my wheel, I didn't try it from top speed and as hard as I could, worrying I'd fall and break vee's wheel. Although, it is said that the current measurement of Gotways is off by a factor of two, so it could have been "only" about 1.5kW (and these are fairly transient spikes occurring only shortly during the braking):

Left-hand scale: speed, voltage, current, temperature. Right-hand scale: power in watts (calculated simply as voltage * current).

EDIT: The largest spikes are due to me pressing down the pedals harder with my legs at the end of the braking.

[DaveThomasPilot]

I didn't notice this thread kept going.  My two cents...

Generally, its more difficult to prevent a motor from charging the battery when the motor is driven externally.  Dumping the energy created by the motor (acting as a generator) into the battery is more convenient from an electrical design standpoint.

Under no load, the motor will generate a voltage that's proportional to its RPM (Kv/winding constant).   If there's not a current path to the battery, the motor will basically "free wheel" and offer no braking.  The motor develops a voltage, but no power is generated.  Power is voltage * current.  Current is zero, so power is zero.

If you try turning a generator under no load (open terminals) you'll find it easy to spin.  When you allow current to flow at the generator terminals (put a resistor or load across the terminals) you'll find it harder to turn.  The higher the load (current) the harder it will be to spin.   This is the "braking".

Braking is eliminated if there's no current path from the motor to the battery when the motor voltage is higher than the battery voltage.

But, this requires that the switches between the battery and motor (speed control) can block current flow in the "reverse" direction.   FETS don't do this--they have an intrinsic diode that always conducts when the source is more positive than the drain.

Bad things can happen when the intrinsic diode conducts high current, so generally a discrete diode is place in parallel with the FET if current flow in the reverse direction is desired, eg, for "regenerative braking".  These diodes must block the full battery voltage (plus inductive spkes) in one direction and carry the full motor current when conducting.  So, they add cost and take up board space.

So, I'd say if the EUC is braking, the energy is going into the battery.  Diodes could be included in the design to prevent this, but then you'd just get freewheeling with no braking.

 

 

 

 

 

 

 

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[chriscalandro]

Yes, if you put all you've got into stopping you will see a big spike. I was talking about normal circumstances and regeneration. In normal riding, after kepping you balanced. Net power return isn't that high unless you are on a pretty good size hill.

 

This discussion is about overcharging the battery, and I think you will find you used more of the battery getting to that speed, than your quick slow down produced. So in terms of this discussion this example shows a total net loss rather than a gain. 

 

If you want to show why a wheel blows up after an abrubt stop, 50 amps will do it for sure. But that doesn't mean you over charged the battery. 

[DaveThomasPilot]

Still thinking about the braking and where the current could go...

Fully charge Lithium batteries won't accept much current.  They're basically open circuit.  So, even if a motor is directly (diode) connected to the battery, the voltage would not be clamped.

So, for braking to work, there would need to be some other path for the current flow.  The overvoltage protection for the battery would need to be a shunt, or clamping type of protection versus a series, or blocking type of protection.

Thinking about this for the first time, so I may be wrong.  But, if the battery is fully charged and you still experience braking, it isn't  going into the battery.  So, there must be some other current path that dissipates the energy.

Has anybody experienced braking on a full battery?

[chriscalandro]

Exactly correct. A battery will pull a load with a direct relationship to it's charge. I mentioned a car battery earlier. 

 

A dead battery will have a large load on a charging circuit. A charged battery will have no load. (My big spark vs little spark analogy)

 

That is why when you plug your ninebot into the wall, you can get it to charge half way quickly while the other half takes a much longer time. The more charged the battery, the less charge it takes in. I hope that makes sense. 

 

As for where does the power go,

A circuit is only created and electrons will only flow when there is a load. No load, no power, no current.   If you have a motor acting as a generator when power isn't needed no power is generated. 

 

What this means for slowing down going down hill

- the motor runs in reverse. This is where it gets really math-ie. 

The higher the load on the motor, the more current it takes to make you go, or inversely, slow you down. The more the motor controller controls the motor to slow you, the higher the load on the motor, the more power you generate, but this can only generate current if the battery draws a load. If the battery is full, the motor WILL actually just run backwards at a rate that evens out what the battery draws (or doesn't draw) with a ultimate net power generated of 0. 

 

Power generated - power called from the battery - power used = 0

 

If the battery is charged, the motor will generate power backwards evening out total net power to 0. 

 

This is confusing, but if makes sense. 

 

And I'll try it one more way. 

 

If this were just a generator and there was no load, the generator would freewheel and no power would be generated. The higher load the harder it is to turn the generator, the more power is produced PROPORTIONALLY TO THE LOAD.

 

Because you cannot have this freewheeling  in our euc example, the motor will be actively driven backward. The load of that, will be compensated by the power generated from the motor, because now that the motor is driving backward there is a load and the net (power produced vs power used) is still 0. The battery has not been charged and no extra power has been generated. 

 

When it comes to the above graph, it doesn't show what gets put into the battery. It only shows what is going on before the battery. Between what the battery pulls in and what the motor needs to do to keep you up AND slow you down, it took 3kw total. That three kilowats are not what went into the battery, but what it took to do the task. 

 

Between the motor and the battery 3kw were used. 

[esaj]

1 hour ago, chriscalandro said:

 

Pardon me, if I ask stupid questions or explain things wrong, I'm just trying to understand how these things work

Quote

Exactly correct. A battery will pull a load with a direct relationship to it's charge. I mentioned a car battery earlier. 

A dead battery will have a large load on a charging circuit. A charged battery will have no load. (My big spark vs little spark analogy)

I've understood that this is created by the potential (voltage) difference of the two sources (motor as generator and battery), if the voltage is the same, no current will flow in either direction (and the motor freewheels), when the back-EMF (voltage) of the motor is larger, the current will flow towards the battery (and slow down the motor at the same time), and when battery voltage is larger, it will drive the motor until an equilibrium is reached (or the motor runs out of torque?). Could be totally wrong though

 

Quote

That is why when you plug your ninebot into the wall, you can get it to charge half way quickly while the other half takes a much longer time. The more charged the battery, the less charge it takes in. I hope that makes sense. 

From what I've understood, the charger has constant current/constant voltage (CC/CV) -modes, first it uses constant current by keeping the charger voltage slightly above the battery voltage to control the maximum current, and once the "full battery" voltage is reached, it switches to constant voltage mode (well, basically just stops raising the voltage?), and the current then slowly becomes smaller as the battery charges up to said voltage. Once the voltage difference is 0V, no more current flows, similarly as above with the generator/battery. So the "slower half" is due to smaller current, ie. the charge builds up slower?

 

Quote

As for where does the power go,

A circuit is only created and electrons will only flow when there is a load. No load, no power, no current.   If you have a motor acting as a generator when power isn't needed no power is generated. 

But wouldn't the motor then freewheel if no current flows? I think this could occur if the low-sides of the half-bridges are closed, even if the diodes inside the high-side mosfets conduct, as there's no path to "ground" (battery negative side).

Quote

What this means for slowing down going down hill

- the motor runs in reverse. This is where it gets really math-ie. 

The higher the load on the motor, the more current it takes to make you go, or inversely, slow you down. The more the motor controller controls the motor to slow you, the higher the load on the motor, the more power you generate, but this can only generate current if the battery draws a load. If the battery is full, the motor WILL actually just run backwards at a rate that evens out what the battery draws (or doesn't draw) with a ultimate net power generated of 0. 

Are you saying that the motor is actually driven in reverse to slow it down, ie. using the battery power, instead of regenerative braking? Wouldn't this still cause heat build up somewhere (in the motor or bridges)? Or just that it stays around the "equilibrium", if it's running too fast, a slight current runs towards the battery, until it slows enough for the voltages to match, and then when it's running too slow, the battery gives a little bit of current for it to catch up?

I've understood that this is how the balancing in general occurs (the motor is constantly "hovering" around the equilibrium state, and PWM duty cycle is changed to alter the voltage that the motor "sees", as the gyroscopes/accelerometers detect a change from the "0-degree" -position)?

 

Quote

Power generated - power called from the battery - power used = 0

If the battery is charged, the motor will generate power backwards evening out total net power to 0. 

This is confusing, but if makes sense. 

I must say I'm confused, from what I've understood this is is the free-wheeling state where there's no torque in the motor, ie. it will not brake or accelerate.

 

2 hours ago, DaveThomasPilot said:

Still thinking about the braking and where the current could go...

Fully charge Lithium batteries won't accept much current.  They're basically open circuit.  So, even if a motor is directly (diode) connected to the battery, the voltage would not be clamped.

So, for braking to work, there would need to be some other path for the current flow.  The overvoltage protection for the battery would need to be a shunt, or clamping type of protection versus a series, or blocking type of protection.

Thinking about this for the first time, so I may be wrong.  But, if the battery is fully charged and you still experience braking, it isn't  going into the battery.  So, there must be some other current path that dissipates the energy.

Has anybody experienced braking on a full battery?

Like I mentioned before, I did experience a (probably) overcharge cut-out (the BMS cut off power) during braking downhill on full batteries (straight from charger, ridden for about 100m before starting to go downhill). I got about 5 meters downhill before it cut out under me. The reason for this seems to be that the BMSs in my custom packs were the type where charging and discharging occurs over the same wires. I've done so hundreds of times with the original batteries (straight of the charger, downhill for a few hundred meters), but they have different type of BMS, with separate P- / C- -wires for charging and discharging, so it would appear that the original batteries have the overcharge-protection only on the charging-side, and the Firewheel mainboard happily lets the batteries overcharge via the discharge-side during braking. (And yeah, I've now got new BMSs for the custom-packs, that have separate charging and discharging points ).

From what I've gathered, there (probably) are transient voltage suppression diodes ("avalanche" diodes) inside the motor, but they would seem only be there to prevent the inductive kickbacks from going into the half-bridges (the transient voltages could be easily hundreds of volts and would strike through the mosfets?), and probably not to limit the voltage towards the batteries. I bought some of those diodes for testing (12V / 1.5kW + 75V / 1.5kW), but haven't tried them yet. I did measure with my cheapo-DIY-oscilloscope that I get over 60V transient spikes from a small  12V (nominal) 3-phase motor just by flicking it fast with my fingers.

 

2 hours ago, DaveThomasPilot said:

I didn't notice this thread kept going.  My two cents...

Generally, its more difficult to prevent a motor from charging the battery when the motor is driven externally.  Dumping the energy created by the motor (acting as a generator) into the battery is more convenient from an electrical design standpoint.

Under no load, the motor will generate a voltage that's proportional to its RPM (Kv/winding constant).   If there's not a current path to the battery, the motor will basically "free wheel" and offer no braking.  The motor develops a voltage, but no power is generated.  Power is voltage * current.  Current is zero, so power is zero.

If you try turning a generator under no load (open terminals) you'll find it easy to spin.  When you allow current to flow at the generator terminals (put a resistor or load across the terminals) you'll find it harder to turn.  The higher the load (current) the harder it will be to spin.   This is the "braking".

Braking is eliminated if there's no current path from the motor to the battery when the motor voltage is higher than the battery voltage.

This is my understanding so far also also.

Quote

But, this requires that the switches between the battery and motor (speed control) can block current flow in the "reverse" direction.   FETS don't do this--they have an intrinsic diode that always conducts when the source is more positive than the drain.

But if the low-side of any half-bridge isn't open, can this actually happen? Ie. there's no path to the negative-side of the battery, even if the high-side mosfet diode would conduct towards the battery?

Quote

Bad things can happen when the intrinsic diode conducts high current, so generally a discrete diode is place in parallel with the FET if current flow in the reverse direction is desired, eg, for "regenerative braking".  These diodes must block the full battery voltage (plus inductive spkes) in one direction and carry the full motor current when conducting.  So, they add cost and take up board space.

So, I'd say if the EUC is braking, the energy is going into the battery.  Diodes could be included in the design to prevent this, but then you'd just get freewheeling with no braking.

Well, the energy has to go SOMEWHERE, I'm just not sure whether the mainboards allow the overcharge of the battery (it would seem so with the Firewheel at least) or if the energy is burned off as heat somewhere else (or by the battery itself).

 

EDIT: Merged this whole mess into a single message, sorry for quoting "out-of-order"

[DaveThomasPilot]

Just restating--

Unless a separate path (dissipative) path is provided for braking, no braking can occur.  A fully (or even near fully charged) lithium battery is pretty much an open circuit.  So, there's nothing a battery charger can do to steer current into the battery--the voltage would just increase until something bad happened to the battery.

So, either the wheel would just free-wheel when going down hill, or another path must be provided for the current flow.

Something like a snubber network.  A freewheel diode or other clamping circuit.  The voltage must be clamped to something higher (probably much higher) than the battery voltage.  Probably at least 2x the battery voltage. 

Just a few components required, but they would have to be high current and high voltage.  If the EUCs don't have this, they should.

So, my question is, do the EUC's actually brake going down hill with full battery charge, or just free wheel.  I know you said "cut-off", but what exactly happened?  Free wheeling?

[chriscalandro]

There's a lot here and I'm on my phone so I'll do what I saw first. 

 

Your car alternator runs at 14v. Your battery is 12v. Google why that is and you have your answer. 

If the motor needs and power generation are equal, and the battery is charged, the wheel is in fact freewheeling for that period as current does not need to flow anywhere. 

But that is only momentary until the status quo changes. 

[DaveThomasPilot]

The motor won't  "be driven" in the reverse direction when going down hill.  That's what we call braking.

Either zero or some positive current will flow from the motor to somewhere else.  If zero current, no "braking" will occur, aka free-wheeling.  If positive current flows,  braking will occur.  If negative current flows (current into the motor) it will go faster downhill.

 

[esaj]

19 minutes ago, DaveThomasPilot said:

Just restating--

Unless a separate path (dissipative) path is provided for braking, no braking can occur.  A fully (or even near fully charged) lithium battery is pretty much an open circuit.  So, there's nothing a battery charger can do to steer current into the battery--the voltage would just increase until something bad happened to the battery.

So, either the wheel would just free-wheel when going down hill, or another path must be provided for the current flow.

Something like a snubber network.  A freewheel diode or other clamping circuit.  The voltage must be clamped to something higher (probably much higher) than the battery voltage.  Probably at least 2x the battery voltage. 

Just a few components required, but they would have to be high current and high voltage.  If the EUCs don't have this, they should.

I can't tell whether they have this or not. If you can figure it out just by looking at the mainboard, here's a hi-res pictures of the older Firewheel mainboard (didn't embed the images so the thread won't take forever to load for slow connections):

- http://ezbe.underkround.fi/unicycle/fw/mainpcb-frontside.jpg

- http://ezbe.underkround.fi/unicycle/fw/mainpcb-backside.jpg

Quote

So, my question is, do the EUC's actually brake going down hill with full battery charge, or just free wheel.  I know you said "cut-off", but what exactly happened?  Free wheeling?

In the cut-out case, the BMS (momentarily) cut the power, I landed on my feet and the wheel continued downhill doing cartwheels. While it was rolling downhill, I saw that the front- and taillights had gone out. When I got to it, the battery display was still back-lit but showing nothing, so I guess the power was cut momentarily and the firmware crashed (it wasn't playing the normal "Please restart the unicycle"-message it does when it tilts too far and cuts the power) and the motor was "limp". Also the lights are turned on by a momentary switch, and did not react to that. I turned the wheel off and back on, and it started normally.

I then tested it further with full charge and carrying it to the start of the hill, and it would cut the power instantly when rolling it downhill and tilting back (it was showing 67.8V with Charge Doctor when took it off the charger, some minutes later, by the time I got to the top of the hill, the display showed something like 67.3, but I doubt that cheap display is very accurate anyway):

 

 

 

With the original batteries, it would just brake normally (ie. keep my speed steady and not let it rise) on that same hill, fresh off the charger, so full charge, even when I did an "all the way balancing" charge with Charge Doctor, that means waiting until the charge current really drops to 0mA (the charger light turns green somewhere around 200-250mA, if memory serves).

[DaveThomasPilot]

When I get my EUC (hopefully Friday from Jason!), I could fully charge the battery then try driving the wheel with another motor and see what happens.  (does the wheel resist turning or not).

Seems it would have to resist, else no self-balancing could occur at full charge.  Which implies there must be some path for the current when the battery is fully charged.  But, perhaps if the voltage from the motor gets too high, the power circuitry "gives up" some how.

I really can't tell from the picture, but I might be able to deduce what the full power train looks like once I get my King Song.  (Unless taking it apart that far will void the warranty).

 

The big caps may be "the snubber".  A diode from the motor would allow regenerative current to flow and charge the cap.

But once the cap is charged to a voltage that's equal to the motor RPM times the motor Kv, no more current would flow.  So, there would have to be a dissipative network like a resistor to some lower voltage like 0 volts

 

 

 

[Jason McNeil]

A couple months back used a datalogger to capture the current/voltage flow from a full battery pack. Results from that test were that the regen would cause the voltage to momentarily spike above the maximum rated voltage. Clearly this is not ideal, some manufacturers are now recommending to disconnect from the charger before fully topping up the pack, if intending to start out at the top of a hill.  

 

 

[esaj]

5 minutes ago, Jason McNeil said:

A couple months back used a datalogger to capture the current/voltage flow from a full battery pack. Results from that test were that the regen would cause the voltage to momentarily spike above the maximum rated voltage. Clearly this is not ideal, some manufacturers are now recommending to disconnect from the charger before fully topping up the pack, if intending to start out at the top of a hill.  

 

Yeah, this again would suggest that at least a momentary overcharge might occur..(?) Since the motor didn't start free-wheeling (right?), there has to be current to cause torque, and that current has to go somewhere. I don't know if the packs actually "soak" it up even if the voltage is there at the battery wires, or if the BMS does something funky (can it for example burn off the excess somewhere without cutting power)... During a long descent, there's a fairly large amount of power to dissipate though, burning it off in small components doesn't seem likely. Storing it in capacitors and then burning it off slower could work, but the capacitors probably charge to full voltage pretty fast (unless were talking "super/ultra capacitors" with whole farads of capacity  ), so that wouldn't likely help on longer descent either.

Hopefully some day one of our battery/motor drive experts could explain this all to us mere mortals and I could stop best-guessing

[OliverH]

16 minutes ago, Jason McNeil said:

A couple months back used a datalogger to capture the current/voltage flow from a full battery pack. Results from that test were that the regen would cause the voltage to momentarily spike above the maximum rated voltage. Clearly this is not ideal, some manufacturers are now recommending to disconnect from the charger before fully topping up the pack, if intending to start out at the top of a hill.  

 

 

Future generations will have other designs. That's a feature limitation of current EUs. They're not build for safety.

5 minutes ago, esaj said:

Yeah, this again would suggest that at least a momentary overcharge might occur..(?) Since the motor didn't start free-wheeling (right?), there has to be current to cause torque, and that current has to go somewhere.

An IT network guy would say: Route it to device 0

[DaveThomasPilot]

You can't overcharge a lithium battery.  You can damage it by applying a voltage higher than the full charge voltage, but it won't accept current unless a damaging voltage is applied.

Likely the regenerative current charges a capacitor snubber as I mentioned above.  The voltage on the cap would not be applied to the battery--it would be diode isolated.

Here's an explanation of a snubber network:

 

http://siegeelectric.com/products/snubbers.aspx?gclid=COeGm-uyysoCFcQkgQodGzoNtg

Here's a better example, included with an H bridge motor driver:

https://www.google.com/search?q=motor+snubber+network&tbm=isch&tbo=u&source=univ&sa=X&ved=0ahUKEwja6vDpssrKAhUI6CYKHc_tCM8QsAQINA&biw=1333&bih=646#tbm=isch&q=h+bridge+snubber+network&imgrc=Mgco0vO9AqzqxM%3A

 

 

One more, this one specific to regenerative case.

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=7&ved=0ahUKEwirkqbAtMrKAhVGMyYKHXLtBPsQFghHMAY&url=https%3A%2F%2Fwww.ab.com%2Fsupport%2Fabdrives%2Fdocumentation%2Ftechpapers%2FRegenOverview01.pdf&usg=AFQjCNFSc090vIMCmurt8LTQ02DaS0kJPg&cad=rja

Note the useful vocabulary.  The motor is either "motoring" or "regenerative".

<quote>

When the rotor of an induction motor turns slower than the speed set by the applied frequency,
the motor is transforming electrical energy into mechanical energy at the motor shaft. This
process is referred to as ‘motoring’. When the rotor turns faster than the synchronous speed set
by a drive output, the motor is transforming mechanical energy from the motor shaft into electrical
energy. It may be a ramp to stop, a reduction in commanded speed or an overhauling load that
causes the shaft speed to be...

</quote>

 

[OliverH]

13 minutes ago, DaveThomasPilot said:

You can't overcharge a lithium battery.  You can damage it by applying a voltage higher than the full charge voltage, but it won't accept current unless a damaging voltage is applied.

Likely the regenerative current charges a capacitor snubber as I mentioned above.  The voltage on the cap would not be applied to the battery--it would be diode isolated.

Here's an explanation of a snubber network:

 

http://siegeelectric.com/products/snubbers.aspx?gclid=COeGm-uyysoCFcQkgQodGzoNtg

Here's a better example, included with an H bridge motor driver:

https://www.google.com/search?q=motor+snubber+network&tbm=isch&tbo=u&source=univ&sa=X&ved=0ahUKEwja6vDpssrKAhUI6CYKHc_tCM8QsAQINA&biw=1333&bih=646#tbm=isch&q=h+bridge+snubber+network&imgrc=Mgco0vO9AqzqxM%3A

 

What happens when the over charging protection is activated? The battery wouldn't be damaged and what happens to the electrical energy as the motor is in generator mode?

[chriscalandro]

I can't keep up with all these. 

 

When you charge a battery general practice is to apply a slightly higher charging voltage. It's not dangerous. That is how EVERY battery charger works.   Look at your ninebot charger. Tell me what the voltage is. Look at your ninebot battery, tell me what the voltage is. 

 

Look at the graph that was posted earlier. When breaking heavily there was a small voltage gain while the battery charged. 

 

Yes, when you are slowing down or going down hill the pwm will produce a negative motor motion. Whether or not this charges the battery is determined by many factors. But yes. Pwm will be telling the motor to run in reverse. 

 

These are facts. Not opinions. In every case 100% of the time, there is no "where does this extra current go". It either charges the battery or no extra current is generated. 

 

If you have a near dead battery then the battery will be a heavy load. That is where the charge protection circuit comes into play. A dead battery will pull a high load, which must be managed. 

[esaj]

@chriscalandro  @DaveThomasPilot  Since writing a post explaining the entire theory motor driving, regenerative braking etc. would probably be too much work, could either of you suggest a good source or book explaining these things? I've gone over numerous documents about 3-phase motor drives that usually explain the basics, like half-bridges, commutation, hall-sensors/rotary-encoders/whathaveyou, PWM, inductive spiking, back-EMF measuring etc. but are very vague or don't tell anything about handling regenerative/rheostatic/plugging/dynamic/whathaveyou braking, more finer details of building a motor controller, sine-wave drives etc. Scraping the info from multiple sources (that sometimes seem even contradictory with each other) is a pain

[Chriull]

@esaj: Could be that you get more and easier information on the theory if you look for synchronous machine - since that is the "main class" of this kind of motor and exists much longer. Imho.

[DaveThomasPilot]

"These are facts. Not opinions. In every case 100% of the time, there is no "where does this extra current go". It either charges the battery or no extra current is generated.  "

I'm not sure if you're disagreeing withing anything I said or not...

I suspect regenerative current is dissipated in a dissipative snubber network.  A Lion battery won't accept current when it's fully charged.  It's an open circuit at full charge.  That's why the most common charge termination technique is constant voltage until charge current drops below a threshold.

If there's braking, there's current flow.  So, there would have to be something like a snubber network.

The current "goes" into a cap.  It charges the capacitor in the snubbing network.

The maximum safe (non damaging) charge votlage is higher than the open circuit battery voltage.  Yes, you can usually exceed the maximum voltage without much damage, but the point is no current flows into the battery (unless the voltage is so high it IS damaging the battery.  So, if the EUC has regenerative braking at full battery charge, it has something like the snubber network I described.

 

Here I'm saying more than I know for sure...

The EUC motors are probably DC brushless motors.  At least three phases, I'd think....

The DC means from battery voltage, versus AC line voltage (like 120 VAC).

For purposes of understanding regenerative braking and snubber networks, you can think of be the brushless motor as just having three sets of windings, each requiring it's own snubber network.

Brushless motor drivers need to deal with optimal commutation timing in addition to handling inductive spikes and regenerative current.  So, you'll likely find a lot of information on that if you search on that.

The best reference really depends on your background.  

 

 

Here's what I think is simple explanation:

http://electronics.stackexchange.com/questions/56186/how-can-i-implement-regenerative-braking-of-a-dc-motor

Also, instead of  attempting to return regenerative energy to the battery, the FETs in the bridge could be turned on to dissipate the energy. 

However, this would be a huge amount of power dissipation compared to what they would otherwise need to handle.  Maybe ok for short bursts of braking/balancing, but maybe not ok for a long continuous run.

Then again, maybe the FETs in the bridge are sized big enough (and with enough heat sinking) to handle this.

[esaj]

31 minutes ago, DaveThomasPilot said:

Here I'm saying more than I know for sure...

The EUC motors are probably DC brushless motors.  At least three phases, I'd think....

From what I know, that's right. My understanding is that they're typically 3-phase outrunner "direct-drive" hub-motors, although some manufacturers (like Rockwheel) have also used geared motors. There are tear down pictures of some of the motors around the forum and elsewhere, for example: http://aijaa.com/Qh533Z

Quote

The DC means from battery voltage, versus AC line voltage (like 120 VAC).

For purposes of understanding regenerative braking and snubber networks, you can think of be the brushless motor as just having three sets of windings, each requiring it's own snubber network.

Brushless motor drivers need to deal with optimal commutation timing in addition to handling inductive spikes and regenerative current.  So, you'll likely find a lot of information on that if you search on that.

The best reference really depends on your background.  

I have little background in electronics, I went to a school around the turn of the millenium that had lots of electrics (AC) and electronics courses, but wasn't that interested in the subject back then, and while I did ok in the courses, I have since forgotten pretty much all I knew back then (which probably wasn't that much anyway ).Now I've started to "re-learn" electronics, but I guess you could say I'm totally amateur or "hobbyist" at best.  My work background is in software.

So far, I've built (based on e-bike schematics) a simple 3-phase motor controller using an Arduino to handle the hall-sensor reading & commutation PWM (no back-EMF -measuring):

 

 

 

 

In the first video, the motor did not yet start on it's own, so I had to help it a little with my finger, in the second it already runs on it's own (constant pwm duty cycle). Next part would be speed control by modifying the duty cycle and getting better power source (those 9V batteries don't seem to cut it, the voltage sags to around 5V when the current goes above 1A ). Also I recently (finally) got a cheap "DIY" (means solder it yourself) oscilloscope so I can finally see better what's going on (I don't know if the problem is too large or too little capacitor in the high-side charge pump, but my guess would be that the voltage doesn't stay high enough for long enough to keep it conducting). That video is from several weeks back, and I must admit I don't remember what I even did to get it turning on it's own, probably changed the capacitors to another value.

This is a "learning project" to understand how the motor controllers work and how they're built. Preferably I'll go with discrete components as much as I can instead of something like mosfet drive ICs, to better understand things.

In the longer run, I'd like to be able to build a self-balancing robot using those 3-phase motors, so far I've started it with simple brushed motors & dual H-bridge drive:

 

Yeah, it's really not working that well yet, plus in that video I still had a bug in the code where the motors would stop in certain angles   I don't have a video of the "next" proto yet, that uses adjustable PID-controls.

Currently, I'm planning on building an adjustable bench power supply (yeah, I'm cheap) from an old computer PSU with multiple adjustable outputs (0...12V, >12V with boost converters) to help with the power requirements (plus I've ordered some higher C-rate NiMH-batteries) and using varying voltages for all sorts of projects. I've been pestering @zlymex and @Cranium about my overcurrent protection circuit design, and finally have it pretty much at the point where I can build it.

In a really, really long run, I'd like to build an entire EUC, but that might never happen... or at least I'm not sure if I'd be crazy enough to try to ride it myself

So far on the wheel-front, I've built a custom frame and used Firewheel mainboard + motor and the custom batteries to ride around with it:

 

 

 

The current problem with this project is that I believe the mainboard is too far from the axle, and the accelerometer in the gyroscope/accelerometer-chip gets too high values for what's it programmed for. The result is that when I hit a bump riding in the woods, and the wheel leans forwards, it really RIPS itself back upright (huge acceleration), and it's downright scary   That project is currently on hold, as the garage is needed for the car until spring

 

Sorry for the image/video overload

[chriscalandro]

Esaj to comment on your project,

The higher the gyro is on a fulcrum, the farther it needs to travel to register the same change as if it were lower. 

 

The gyro itself should be as close to the bottom as possible to increase resolution. 

[DaveThomasPilot]

 

Nice project, ESAC!  Seems like you're beyond the basics already.

I'm surprised no one mentioned this thread (I found it last night):

 

 

[chriscalandro]

The thing about that thread and all that math is that it skips the sime point that a generator will only supply current when there is a load. No load, no current, no charging. 

 

If the battery is charged and is not pulling a load on the circuit, essentially, that is no longer a circuit, and no current will be generated meaning the battery will not be charged. 

 

The only way to overcharge would be to use a higher voltage. Which can happen with big spikes, but a simple DC regulator can cover.